Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{4y - 28}{4y^2 - 28y + 40} \times \dfrac{y^2 - 8y + 15}{-3y + 9} $
Answer: First factor out any common factors. $r = \dfrac{4(y - 7)}{4(y^2 - 7y + 10)} \times \dfrac{y^2 - 8y + 15}{-3(y - 3)} $ Then factor the quadratic expressions. $r = \dfrac {4(y - 7)} {4(y - 5)(y - 2)} \times \dfrac {(y - 5)(y - 3)} {-3(y - 3)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {4(y - 7) \times (y - 5)(y - 3) } { 4(y - 5)(y - 2) \times -3(y - 3)} $ $r = \dfrac {4(y - 5)(y - 3)(y - 7)} {-12(y - 5)(y - 2)(y - 3)} $ Notice that $(y - 5)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {4\cancel{(y - 5)}(y - 3)(y - 7)} {-12\cancel{(y - 5)}(y - 2)(y - 3)} $ We are dividing by $y - 5$ , so $y - 5 \neq 0$ Therefore, $y \neq 5$ $r = \dfrac {4\cancel{(y - 5)}\cancel{(y - 3)}(y - 7)} {-12\cancel{(y - 5)}(y - 2)\cancel{(y - 3)}} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $r = \dfrac {4(y - 7)} {-12(y - 2)} $ $ r = \dfrac{-(y - 7)}{3(y - 2)}; y \neq 5; y \neq 3 $